Ejecting electron from metal Physics Please Help?

Question

The minimum energy required to remove an electron from a metal is 2.7 eV. What is the longest wavelength photon that can eject an electron from this metal?

Answer 1

we can use the photoelectric effect to calculate the longest wavelength:

the formula of the photoelectric effect is:



Kmax = h * f - φ

where:

Kmax is the maximum kinetic energy of the ejected electron

h is planks constant

f is the frequency of the incident photon and

φ is the work function, which is the minimum energy required to remove an electron from the surface of any given metal, so:



further more:

φ = h * f0

where f0 is the threshold frequency, so:



Kmax = h * f - h * f0

Kmax = h * (f - f0)



so as you can see φ = 2.7 eV , because the minimum energy required to remove an electron from the metal is 2.7 eV, so:



φ = 4.325 * 10^-19 J

h * f0 = 4.325 * 10^-19

f0 = 6.528 * 10^14 Hz



because the electron kinetic energy has to be positive than the minimum f (the frequency of the photons ejecting the electrons) has to be f0, or in other words:

f > f0



since were looking for the longest wavelength, we need the shortest frequency we can which in our case is 6.528 * 10^14 Hz.



the wavelength of the photons in this case is:

the formula for wavelength is:



λ = v / f

where:

λ is the wavelength

v is the velocity of the wave in the medium (in our case its the speed of light, c) and

f is the frequency of the wave , so:



λ = 299 792 458 / (6.528 * 10^14)

λ = 4.592 * 10^-7 meters



the longest wave length is:

λ = 459.240 nanometers



hope i helped and good luck :)

Answer 2

I will give you the method and you can put the numbers in yourself for good practice.



The longest wavelength photon will also be the lowest energy photon, as it will have the smallest frequency and hence lowest energy.

This photon has the same energy as the electron it excites, by energy conservation.

So the lowest energy photon has energy = the lowest energy electron = 2.7eV



We know e=hf where f is the frequency and h is plancks constant

so 2.7eV = hf = hc/(wavelength) as we know c=f(lambda)



=> Wavelength = hc/E=hc/(2.7eV)



Just plug in the numbers, c=3E8 and 1eV=1.602E-19

Answer 3

find the energy of a photon with wavelength 196 nm = 1.96x10^-7 m from E= h c /L where h=6.6x10^(-34 Js c=3 x01^8m/s L=wavelength = 1.96 x 10^-7 m then realize that E(photon) = KE(electron) +Binding Energy Binding Energy = E(photon)-KE(electron) and find the KE in J using the conversion 1eV=1.6x10^-19J